solve this gemotery questions

I think our answers are the same aren't they?  I just like being long winded.  

I solve these and write the code at the same time.  That's probably why I get in such a mess and it takes so long.  But it is more fun that way.  

1/[csc A - cotA] - 1/sinA = 1/sinA  - 1/[csc A + cot A]     get everything in terms of sine and cosine

1/ [(1/sinA - cosA/sinA)] - 1/sinA = 1/sinA - 1/ [(1/sinA + cosA/sinA)]    simplify this

sinA/[1 - cosA] - 1/sinA = 1/sinA - sinA/[1 + cosA]

Multiply the first term on the LHS by (1 + cos A) in the the numerator and the denominator. The denominator becomes (1-cos^A) = sin^2A.......the same sort of thing is also done to the last term on the RHS....

sinA[1 + cosA]/sin^2A - 1/sinA = 1/sinA - sinA[1-cosA]/sin^2A

Get common denominators - (sin^2A) - on both sides

sinA[1+cosA]/sin^2A -sinA/sin^2A  = sinA/sin^2A - sinA[1 - cosA]/sin^2a

sinA/sin^2a + cosA/sin^A - sinA/sin^2A = sinA/sin^2A - sinA/sin^2A + cosA/sin^2A

cosA/sin^2A    = cosA/sin^2A

Sorry Chris, I can't resist.    

Your will be amused to know that I got in a mess and it took forever.   lol

$${\frac{{\mathtt{1}}}{\left({\mathtt{cosecA}}{\mathtt{\,-\,}}{\mathtt{cotA}}\right)}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{sinA}}}} = {\frac{{\mathtt{1}}}{{\mathtt{sinA}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{\left({\mathtt{cosecA}}{\mathtt{\,\small\textbf+\,}}{\mathtt{cotA}}\right)}}$$

$$\begin{array}{rll}
LHS&=&\frac{sinA}{sinA(cosecA-cotA)}-\frac{(cosecA-cotA)}{sinA(cosecA-cotA)}\\\\
&=&\frac{sinA-(cosecA-cotA)}{sinA(cosecA-cotA)}\\\\
&=&\frac{sinA-cosecA+cotA}{sinA(\frac{1}{sinA}-\frac{cosA}{sinA})}\\\\
&=&\frac{sinA-cosecA+cotA}{1-cosA}\\\\
&=&\frac{(sinA-cosecA+cotA)sinA}{(1-cosA)sinA}\\\\
&=&\frac{sin^2A-1+cosA}{(1-cosA)sinA}\\\\
&=&\frac{1-cos^2A-1+cosA}{(1-cosA)sinA}\\\\
&=&\frac{cosA(1-cosA)}{(1-cosA)sinA}\\\\
&=&\frac{cosA}{sinA}\\\\
&=& CotA\\\\\\
RHS&=&\frac{(cosecA+cotA)}{sinA(cosecA+cotA)}-\frac{sinA}{sinA(cosecA+cotA)}\\\\
&=&\frac{cosecA+cotA-sinA}{1+cosA}\\\\
&=&\frac{(1+cosA-sin^2A)}{(1+cosA)sinA}\\\\
&=&\frac{(1+cosA-(1-cos^2A)}{(1+cosA)sinA}\\\\
&=&\frac{cosA(1+cosA)}{(1+cosA)sinA}\\\\
&=&\frac{cosA}{sinA}\\\\
&=& cotA\\\\
&=& LHS \qquad QED\\\\







\end{array}$$

That's one thing about these identities......there are many approaches, sometimes......